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\title{第二章：随机变量及其分布}
\author{MSS ET AL}
\date{2018年10月}

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\begin{frame}{内容提要  }

\begin{itemize}

\item  随机变量的概念、分布函数

\item  随机变量的密度函数、分布列

\item  离散型：二项分布、泊松分布

\item   连续型：正态分布、指数分布、均匀分布

\item   随机变量的数学期望与方差

\item   随机变量的函数

\end{itemize}

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\begin{frame}{本章习题}

\begin{itemize}

\item   (2.1) 1,5,8,9,10,12,16,17.

\item   (2.2) 1,4,10,12,15,18,19.

\item   (2.3) 1,3,6,8,10,12.

\item   (2.4) 1,4,7,9,13,15.

\item   (2.5) 1,3,4,6,9,11,17,19,23,29.

\item   (2.6) 1,3,4,5,7,8,12.

\end{itemize}

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问：怎么理解{\color{red}随机变量}这个概念？检查三个产品，考察其中次品的数目，如何描述这个随机变量？


答：理解随机变量的两个关键：
\begin{itemize}
\item 写出样本空间 $\Omega$
\item 写出对应关系 $X:\Omega\to\mathbb{R}$
\end{itemize}

%两种情形：\\ (a)样本点是数字，(b)样本点不是数字。




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问：什么是{\color{red}随机变量的分布函数}？向半径为r的圆内随机抛一点。 记$X$ =该点到圆心的距离。(a) 求$X$的分布函数$F(x)$. (b) 求概率$P(X > 2r/3)$.


\begin{itemize}
\item 分布函数记录了随机变量的概率分布情况。
\item $F(x)=P(X\le x), -\infty<x<+\infty$. 
\item 当 $0\le x\le r$ 时，$F(x)=\frac{\pi x^2}{\pi r^2}$.
\end{itemize}


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问：证明{\color{red}分布函数}的基本性质：(a)单调递增，(b)从0增加到1，(c)右连续。


\begin{itemize}
\item 当 $x_1<x_2$ 时有 $F(x_1)\le F(x_2)$.
\item 由 $P(\varnothing)=0$, $P(\Omega)=1$ 得。
\item 当 $x\to a+$ 时有 $F(x)\to F(a)$.
\item cdf = cumulative {\color{red}distribution function}
\end{itemize}



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问：掷两颗骰子，求随机变量的{\color{red}分布列(律)}。
\begin{enumerate}
\item $X$ =‘两个点数之和’，
\item $Y$ =‘出现6的次数’，
\item $Z$ =‘两个点数中较大者’。
\end{enumerate}


答：先写出随机变量的可能取值。再计算各个取值的概率。写成表格，称为{\color{red} distribution}.


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问：已知随机变量 $X$ 的分布列
\begin{table}[ht]\centering
\begin{tabular}{r|c|c|c}
$X$ & $-1$ & $2$ & $3$ \\
\hline
概率  & $0.25$ & $0.5$ & $0.25$ \\
\end{tabular}
\end{table}
\begin{enumerate}
\item 求概率$P(X\le 0.5)$ 和 $P(1.5<X\le 2.5)$.
\item 求分布函数$F(x)$.
\end{enumerate}
%写出分布列的基本性质：非负性，正则性。


答：$P(X\le 0.5)=0.25$. $F(x)$是阶梯函数。


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问：一车沿街行驶，依次经过三个红绿灯。记 $X$=‘首次遇到红灯前已经通过的路口数目’，求 $X$ 的分布列。设红绿灯的时间间隔相同。


答：$X$ 的可能取值为 $0,1,2,3$, 分布列为下表。
\begin{table}[ht]\centering
\begin{tabular}{p{1cm}|p{1cm}|p{1cm}|p{1cm}|p{1cm}}
$X$ & $0$ & $1$ & $2$ & $3$ \\
\hline
概率  & $1/2$  & $1/4$  & $1/8$  & $1/8$ \\
\end{tabular}
\end{table}


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问：连续随机变量与离散随机变量的区别有哪些？解释概率密度函数的形成过程、不同形状。



\begin{itemize}
\item 连续型的分布函数是连续的，离散型的分布函数是阶梯的。
\item 连续型的概率在单点都是零，离散型的概率就集中在若干单点上。
\item 连续型有密度函数，离散型有分布列。
%\item 连续随机变量的概率密度函数在‘几乎处处相等’的意义上来定义的。
\end{itemize}


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问：什么是{\color{red}概率密度函数}？证明其基本性质。


\begin{itemize}
\item 若分布函数 $F(x)$ 可以写成某函数 $p(x)$ 的不定积分，则称 $p(x)$ 是概率密度函数。
\item 由概率密度函数可以很方便地计算概率。
\item $P(a\le X\le b) = \int_a^b p(x)dx$.
\item pdf = probability density function.
\end{itemize}


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问：向区间 $(0,a)$ 上任意投点，用 $X$ 表示这个点的坐标。设这个点落在 $(0,a)$ 中任意一个小区间的概率与这个小区间的长度成正比，而与小区间的位置无关。 求 $X$ 的分布函数和密度函数。


答：先求分布函数。分情况讨论。\\
当 $x\le 0$ 时：$F(x)=P(X\le x)=P(\varnothing)=0$,\\
当 $0< x< a$ 时：$F(x)=x/a$,\\
当 $x\ge a$ 时：$F(x)=1$.

再由 $p(x)=F'(x)$ 求概率密度函数。




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问：设某种型号的电子元件的寿命 $X$ 有以下概率密度函数：
\begin{eqnarray*}
p(x)=\left\{\begin{array}{ll}
1000/x^2, & x>1000 \\
0, & x\le 1000.
\end{array}\right.
\end{eqnarray*}
现有一大批这种元件，其工作相互独立，求：\\
(1) 任取1只其寿命大于1500小时的概率。\\
(2) 任取4只其寿命都大于1500小时的概率。\\
(3) 任取4只至少1只其寿命大于1500小时的概率。\\
(4) 若已知某元件寿命大于1500小时, 问其寿命大于 2000 小时的概率。

%
%答：

%
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%问：概率密度函数与概率分布列的异同。
%%答：


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问：设某随机变量有下述分布函数，解释该随机变量既不是离散型的也不是连续型的：
\begin{eqnarray*}
F(x)=\left\{\begin{array}{ll}
0, & x<0, \\
(1+x)/2, & 0\le x< 1,\\
1, & x\ge 1.
\end{array}\right.
\end{eqnarray*}


\begin{itemize}
\item 连续型随机变量的分布函数是连续函数。
\item 离散型随机变量的分布函数是分段常数函数。
\end{itemize}


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问：甲乙各出50元。设先赢三局者胜。现甲赢两局乙赢一局。求如何分赌本。


\begin{itemize}
\item 分别计算甲和乙先赢三局的概率。
\item 若没有‘先赢三局者胜’的假设。
\item 若假设为‘先赢五局者胜’。
\end{itemize}


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问：什么是{\color{red}数学期望}？离散型和连续型随机变量的数学期望是怎么计算的？


\begin{itemize}
\item 数学期望是以概率为权重的某种平均值。
\item 离散型：$E(x)=x_1p_1+x_2p_2+\cdots$
\item 连续型：$E(X)=\int_{-\infty}^{+\infty} xp(x)dx$
\end{itemize}


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问：同时抛五颗骰子，求出现6的个数的数学期望。


\begin{itemize}
\item 出现6的个数可能为0,1,2,3,4,5.
\item 分别求出概率 $p_0,p_1,p_2,p_3,p_4,p_5$.
\item 按公式计算。
\item 另解：$E(X_1+\cdots+X_5)=E(X_1)+\cdots+E(X_5)$.
\end{itemize}


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问：(1) 求均匀分布的随机变量的数学期望。\\ (2) 求柯西分布的随机变量的数学期望。


\begin{itemize}
\item $E(X)=\int_\mathbb{R} xp(x)dx=\int_a^b \frac{x}{b-a}dx=\frac{b+a}{2}$.
\item %柯西分布的密度函数看起来中心位置是原点，但
积分不是绝对收敛，所以数学期望不存在：
$E(|X|)=\int_\mathbb{R} |x|p(x)dx=\int_\mathbb{R} \frac{x}{\pi(x^2+1)}dx=\infty$.
\end{itemize}


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问：已知随机变量 $X$ 的分布列，求随机变量 $X^2$ 的分布列，和 $X^2$ 的数学期望。
\begin{table}[ht]\centering
\begin{tabular}{c|ccccc}
$X$ & $-2$ & $-1$ & 0 & 1 & 2 \\
\hline
概率 & 0.2 & 0.1 & 0.1 & 0.3 & 0.3 \\
\end{tabular}
\end{table}


答：$X^2$ 的可能取值为 $0,1,4$. 概率分别为 $0.1$, $0.4$, $0.5$. 写成表格就是 $X^2$ 的分布列。
取值和取值概率相乘相加就得到数学期望。



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问：证明随机变量的数学期望有下述性质：
\begin{enumerate}
\item {\color{red}随机变量的函数}的数学期望计算公式
\item 常数的数学期望
\item 求数学期望保持加法和数乘
\end{enumerate}


$E[g(X)]=\sum g(x_i)p(x_i)$.

$E[g(X)]=\int_\mathbb{R} g(x)p(x)dx$.


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问：设原料的市场需求是一个服从 (300,500) 上均匀分布的随机变量。 设每售出1吨可获利1.5千元， 每积压1吨要损失0.5千元。 问进货多少可以使平均收益最大？


答：设进货 $a$ 吨，作为参数。设需求是 $X$ 吨，收益是 $Y$ 千元，则 $X$ 和 $Y$ 都是随机变量，而且 $Y$ 是 $X$ 的函数。由 $X$ 的分布可以求的 $Y$ 的分布。


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问：随机变量的{\color{red}方差和标准差}是用来描述什么的？写出它们的计算公式。%（分离散型和连续型随机变量）



\begin{itemize}
\item 是描述随机变量取值的离散程度的。
\item $Var(X)=E[(X-\mu)^2]$, 这里记 $\mu=E(X)$.
\item Var = Variance = 方差。
\item std = standard deviation = 标准差。
\end{itemize}


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问：考虑随机变量 $X,Y$，其密度函数分别如下，比较其期望与方差。
{\footnotesize
\begin{eqnarray*}
p_X(x) &=& \left\{\begin{array}{ll} 
1+x,& -1\le x<0\\
1-x,& 0\le x<1\\
0, & \text{其它} x.
\end{array}\right.\\
p_Y(x) &=& \left\{\begin{array}{ll}
1/2, & -1<x<1\\
0, & \text{其它} x.
\end{array}\right.
%p_Z(x) &=& \left\{\begin{array}{ll} 
%-x,& -1\le x<0\\
%x,& 0\le x<1\\
%0, & \text{其它} x.
%\end{array}\right.
\end{eqnarray*}
}


答：由密度函数的图像直观可知，\\
$E(X)=E(Y)$, $Var(X)<Var(Y)$.
%$Var(X)=E(X^2)-E(X)^2$.


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问：设随机变量 $X$ 与 $Y$ 分别定义如下。
$X$= ‘投资房地产的收益’，
$Y$= ‘投资商业的收益’，
已知它们的分布列如下表，比较其期望与方差。
\begin{table}[ht]\centering
\begin{tabular}{c|ccc||c|ccc}
$X$ & 11 & 3 & $-3$ & $Y$ & 6 & 4 & $-1$\\
\hline
概率 & 0.2 & 0.7 & 0.1 &概率 & 0.2 & 0.7 & 0.1
\end{tabular}
\end{table}


答：$E(X)>E(Y)$,\,\, $Var(X)>Var(Y)$.


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问： 证明方差的性质：
\begin{enumerate}
\item $Var(X)=E(X^2)-E(X)^2$.
\item $Var(C)=0$.
\item $Var(aX+b)=a^2Var(X)$.
\end{enumerate}


答：


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问：投掷一颗骰子。设 $X$ 为出现的点数。求 $X$ 的期望和方差。


\begin{itemize}
\item 写出分布列。
\item 计算 $E(X)$ 和 $E(X^2)$.
\item 计算 $Var(X)$.
\end{itemize}


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%A\thepage 问：
证明{\color{red}切比雪夫不等式}：设随机变量 $X$ 的数学期望和方差都存在，则对任意正数 $\varepsilon$, 有
\begin{align*}
{}& P\Big{(}|X-E(X)|\ge \varepsilon \Big{)} = \int_{|x-\mu|\ge\varepsilon} p(x)dx \\
\le {}& \int_{|x-\mu|\ge\varepsilon} \frac{(x-\mu)^2}{\epsilon^2}p(x)dx\\
\le {}& \int_{\mathbb{R}} \frac{(x-\mu)^2}{\epsilon^2}p(x)dx \le \frac{Var(X)}{\varepsilon^2}.
\end{align*}



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问：设$X$ = $n$重{\color{red}伯努利试验}中事件 $A$ 发生的次数，设$P(A)=p$. 写出 $X$ 的分布列。


答：记$q=1-p$，
\begin{table}[ht]\centering
\begin{tabular}{|r|c|c|c|c|c|c|c}
\hline
$X$ & 0 & 1 & $\cdots$ & $k$ & $\cdots$ & $n$ \\
\hline
概率  & $q^n$ & $nq^{n-1}p$ & $\cdots$  & $\binom{n}{k}p^kq^{n-k}$ & $\cdots$ & $p^n$ \\
\hline
\end{tabular}
\end{table}

称随机变量 $X$ 服从{\color{red}二项分布}。$X\sim b(n,p)$.


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问：某特效药的有效率为0.8, 今有10人服用，求至少有8人治愈的概率。


答：事件 $A$ = ‘某人服药得治愈’。
设随机变量 $X$ = ‘10人中得治愈的人数’ 。%= ‘10次试验中事件 $A$ 发生的次数’。
则 $X\sim b(n,p)$, 其中 $n=10$, $p=0.8$.
\[ P(X\ge 8) = \sum\limits_{k=8}^{10} \binom{10}{k} (0.8)^k(0.2)^{10-k}. \]



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问：设随机变量 $X\sim b(2,p)$, $Y\sim b(3,p)$, \\
若 $P(X\ge 1)=5/9$, 求 $P(Y\ge 1)$.


\begin{itemize}
\item 由 $P(X\ge 1)=5/9$ 求得参数 $p$.
\item 计算 $P(Y\ge 1)$.%=1-P(Y=0)=$
\end{itemize}


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问：设随机变量 $X$ 服从{\color{red}二项分布} $b(n,p)$, 求 $X$ 的数学期望和方差。

\begin{align*}
E(X) {}& =\sum\limits_{k=0}^{n} k\binom{n}{k} p^k(1-p)^{n-k}=np.\\
E(X^2) {}& =\sum\limits_{k=0}^{n} k^2\binom{n}{k} p^k(1-p)^{n-k}=\\
Var(X) {}& =np(1-p).
\end{align*}



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问：甲乙两人有10局比赛，赢多为胜。设每局甲赢概率为0.6, 乙赢概率为0.4. 设各局比赛相互独立。求最后甲胜、乙胜、甲乙不分胜负的概率。


答：设 $X$ = 甲赢的次数，则 $X\sim b(n,p)$, 其中 $n=10$, $p=0.6$. 
所以甲胜的概率是
\[P(X\ge 6) = \sum\limits_{k=6}^{10} \binom{10}{k} (0.6)^k(0.4)^{10-k}.\]


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问：写出0-1分布 $X\sim b(1,p)$的分布列。


答：0-1分布的可能取值是0或1. 分布列为

\begin{table}[ht]\centering
\begin{tabular}{c|cc}
$X$ & 0 & 1  \\
\hline
概率 & $1-p$ & $p$  \\
\end{tabular}
\end{table}


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问：写出{\color{red}泊松分布} $X\sim P(\lambda)$的分布列，
举出一些符合泊松分布的随机变量$X$的例子。


答：$X=$ 某段时间内某事件发生的次数。
\begin{table}[ht]\centering
\begin{tabular}{|r|c|c|c|c|c|}
\hline
$X$ & 0 & 1 & $\cdots$ & $k$ & $\cdots$  \\
\hline
概率  & $e^{-\lambda}$ & $\lambda e^{-\lambda}$ & $\cdots$  & $\frac{\lambda^k}{k!}e^{-\lambda}$ & $\cdots$ \\
\hline
\end{tabular}
\end{table}

例子：某超市某天的顾客人数。某段时间内在某车站的等车人数。


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问：证明：若$X$服从参数为$\lambda$的泊松分布，则其期望与方差分别为
$E(X) = \lambda$ 和 $Var(X) = \lambda$. 

\begin{align*}
E(X) {}& =\sum\limits_{k=0}^{\infty} k \cdot \frac{\lambda^k}{k!}e^{-\lambda} = \\
E(X^2) {}& =\sum\limits_{k=0}^{\infty} k^2 \cdot \frac{\lambda^k}{k!}e^{-\lambda} = \\
Var(X) {}& =
\end{align*}


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问：某商品的月销售量服从参数为8的泊松分布。问每月进货多少，可以以90\%的概率不会售完。


\begin{itemize}
\item 已知 $X\sim P(8)$, 要找 $n$ 使 $P(X\le n)\ge 0.90$.
\item 由泊松分布，$P(X\le n)= \sum\limits_{k=0}^{n} \frac{\lambda^k}{k!}e^{-\lambda}$. 
\item 查泊松分布表。
\end{itemize}


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问：证明泊松定理：如果 $n$ 较大而 $p$ 较小，则
\begin{eqnarray*}
\binom{n}{k}p^k(1-p)^{n-k} \approx \frac{(np)^k}{k!}e^{-np}.
\end{eqnarray*}


答：

结论：二项分布的极限成为泊松分布。


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问：已知某疾病发生的概率为 0.001,  某单位共 5000 人。 求该病发生人数不超过5人的概率。


答：设 $X$ = 该病发生人数。则 $X\sim b(n,p)$, 其中 $n=5000$, $p=0.001$. 
由泊松定理，$X\overset{\cdot}{\sim} P(\lambda)$, 其中 $\lambda=np=5$.
所求概率约等于
\[ P(X\le 5) \approx \sum\limits_{k=0}^{5} \frac{\lambda^k}{k!}e^{-\lambda} = \text{查表}\]



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问：每台设备发生故障的概率是0.01. 各设备相互独立。 求下述情形下发生故障而不能及时维修的概率：
%\begin{enumerate}
%\item 
(1) 1名维修工负责20台设备。
%\item 
(2) 3名维修工负责90台设备。
%\item 
(3) 10名维修工负责500台设备。
%\end{enumerate}
%\vspace{-0.3cm}


答：设 $X$ = 同时发生故障的设备数目。\\
(1) 已知 $X\sim b(20,0.01)$, 求 $P(X\ge 2)=$\\
(2) 已知 $X\sim b(90,0.01)$, 求 $P(X\ge 4)=$\\
(3) 已知 $X\sim b(500,0.01)$, 求 $P(X\ge 11)=$\\
用泊松近似，然后查表。

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问：写出{\color{red}正态分布}的密度函数。


答：若随机变量 $X$ 的密度函数为
\begin{eqnarray*}
p(x)=\frac{1}{\sqrt{2\pi}\sigma}\exp\left[-\frac{(x-\mu)^2}{2\sigma^2}\right].
\end{eqnarray*}
则称 $X$ 服从均值为 $\mu$, 方差为 $\sigma^2$ 的正态分布，记作 $X\sim N(\mu,\sigma^2)$.


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问：写出{\color{red}标准正态分布}的密度函数。


答：标准正态分布的密度函数为
\begin{eqnarray*}
\varphi(x)=\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{x^2}{2}\right).
\end{eqnarray*}
记作 $X\sim N(0,1)$.


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问：设 $U\sim N(0,1)$, 查表计算下述概率：
\begin{enumerate}
\item $P(U<1.52)$ 和 $P(U>1.52)$ 和 $P(U<-1.52)$.
\item $P(-0.75\le U\le 1.52)$ 和 $P(|U|\le 1.52)$.
\end{enumerate}


答：重点要理解表格上方的公式
\[\Phi(u)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{u} e^{-t^2/2}dt. \]


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问：证明正态分布的期望和方差正好是分布的两个参数：若 $X\sim N(\mu,\sigma^2)$, 则
$E(X)=\mu$ 和 $Var(X)=\sigma^2$.


答：方法一：直接积分计算。可用变量代换。%方法二：转化为标准正态分布。

\begin{itemize}
\item $E(X)=\int_\mathbb{R} xp(x)dx= \int_\mathbb{R} \frac{x}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}} dx=$
\item $E(X^2)=\int_\mathbb{R} x^2p(x)dx=$
\end{itemize}


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问：证明正态分布的标准化定理：

若 $X\sim N(\mu,\sigma)$, 设$U=\frac{X-\mu}{\sigma}$, 则 $U\sim N(0,1)$.


\begin{itemize}
\item 先求 $U$ 的分布函数 $F_U(u)=P(U\le u)=$
\item 再求 $U$ 的密度函数 $p_U(u)=F_U'(u)=$
\item 发现 $U$ 服从标准正态分布。
\end{itemize}


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问：设 $X\sim N(108,3^2)$, 求：\\
(1) 概率 $P(102<X<117)$.\\
(2) 使得 $P(X<a)=0.95$ 的常数 $a$.


\begin{itemize}
\item 转化为标准正态分布。$U=\frac{X-108}{3}\sim N(0,1)$. 
\item $P(102<X<117) = P(-2<U<3)=\Phi(3)-\Phi(-2)=$.
\item $P(X<a)=P(U<\frac{a-108}{3})=\Phi(\,\,)$
\end{itemize}


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%问：设 $X\sim N(\mu,\sigma)$, 求下述概率：
%\begin{enumerate}
%\item $P(|X-\mu|\le \sigma)$.
%\item $P(|X-\mu|\le 2\sigma)$.
%\item $P(|X-\mu|\le 3\sigma)$.
%\end{enumerate}

%
%答：分别是 $\Phi(1)-\Phi(-1)$, $\Phi(2)-\Phi(-2)$ 和 $\Phi(3)-\Phi(-3)$.

%
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问：什么是服从{\color{red}均匀分布}的随机变量？画出其密度函数的图象。


答：若随机变量 $X$ 的密度函数为下述函数，则称 $X$ 服从均匀分布，记为 $X\sim U(a,b)$:
\begin{eqnarray*}
p(x)=\left\{\begin{array}{ll}
\frac{1}{b-a}, & a< x< b \\
0, & \text{其它} x.
\end{array}\right.
\end{eqnarray*}


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问：设随机变量 $X$ 服从 $(0,10)$ 上的均匀分布，现对 $X$ 进行 4 次独立观察，求至少有 3 次观察值大于 6 的概率。


\begin{itemize}
\item $A$=某次观察取值大于6，$P(A)=0.4$.
\item 求4次试验中事件$A$至少发生3次的概率。
\item 二项分布：$\binom{4}{3}(0.4)^3(0.6)^1+\binom{4}{4}(0.4)^4(0.6)^0$.
\end{itemize}


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问：设随机变量 $X$ 服从区间 $(a,b)$ 上的均匀分布， 求 $X$ 的数学期望和方差。


\begin{itemize}
\item 写出均匀分布的密度函数 $p(x)$.
\item 计算一阶矩 $E(X)=\int_\mathbb{R} xp(x)dx=$
\item 计算二阶矩 $E(X^2)=\int_\mathbb{R} x^2p(x)dx=$
\item $Var(X)=E(X^2)-E(X)^2=$
\end{itemize}


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问：什么是服从{\color{red}指数分布}的随机变量？画出其密度函数的图象。


答：若随机变量 $X$ 的密度函数如下，则称 $X$ 服从均值为 $1/\lambda$ 的指数分布， 
记作 $X\sim Exp(\lambda)$:
\begin{eqnarray*}
p(x)=\left\{\begin{array}{ll}
\lambda e^{-\lambda x}, & x\ge 0 \\
0, & x<0.
\end{array}\right.
\end{eqnarray*}


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问：证明：若某设备在任何时段 $[0,t]$ 的发生故障的次数 $N(t)$ 服从参数为 $\lambda t$ 的泊松分布，则相继两次故障之间的时间间隔 $T$ 服从均值为 $1/\lambda$ 的指数分布。


\begin{itemize}
\item 先求分布函数 $F(t)=P(T\le t)=P[N(t)\ge 1]$.
\item $F(t)=1-P[N(t)=0]=1-e^{\lambda t}$.
\item 这正好是指数分布的分布函数。
\end{itemize}


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问：计算服从指数分布的随机变量的期望和方差：
设随机变量 $X$ 服从参数为 $\lambda$ 的指数分布，则有\\
$E(X)=1/\lambda$ 和 $Var(X)=1/\lambda^2$.


\begin{itemize}
\item 写出指数分布的密度函数 $p(x)$.
\item 计算一阶矩 $E(X)=\int_\mathbb{R} xp(x)dx=$
\item 计算二阶矩 $E(X^2)=\int_\mathbb{R} x^2p(x)dx=$
\item $Var(X)=E(X^2)-E(X)^2=$
\end{itemize}


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问：证明服从指数分布的随机变量没有记忆
$$P(X>s+t|X>s)=P(X>t).$$


\begin{itemize}
\item 写出条件概率的计算公式 $P(A|B)=\frac{P(AB)}{P(B)}$.
\item 这里 $A=(X>s+t)$, $B=(X>t)$.
\item 用指数分布的密度函数计算等式两边的概率。
\end{itemize}


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问：设 $X$ 是一个随机变量， $y=g(x)$ 是一个实数到实数的函数，则 $Y=g(X)$ 也是一个随机变量。如何理解定义这个随机变量的样本空间和对应法则。


答：
\begin{align*}
{\Omega} & \overset{X}{\longrightarrow} {\mathbb{R}}  \overset{g}{\longrightarrow}  {\mathbb{R}} \\
{\omega} & \mapsto  {X(\omega) } \mapsto  {g(X(\omega))}
\end{align*}


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问：设随机变量 $X$ 的分布列如下，求随机变量 $Y=X^2+X$ 的分布列：
\begin{table}[ht]\centering
\begin{tabular}{c|ccccc}
$X$ & $-2$ & $-1$ & 0 & 1 & 2 \\
\hline
概率 & 0.2 & 0.1 & 0.1 & 0.3 & 0.3 \\
\end{tabular}
\end{table}


\begin{itemize}
\item 计算 $Y$ 的可能取值，和取这些值的概率。
\item 写成表格形式。
\end{itemize}


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%问：正态分布的一些练习：
%\begin{enumerate}
%\item 设 $X\sim N(\mu,\sigma^2)$, 求 $Y=aX+b$ 的分布。
%\item 设 $X\sim N(10,2^2)$, 求 $Y=3X+5$ 的分布。
%\item 设 $X\sim N(0,2^2)$, 求 $Y=-X$ 的分布。
%\end{enumerate}

%
%答：1. $Y\sim N(a\mu+b, a^2\sigma^2)$.

%
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问：设 $X\sim N(0,1)$, 求 $Y=X^2$ 的分布。


\begin{itemize}
\item 求‘分布’一般指的是求分布函数。\\（或求密度函数、分布列）
\item 若与有名称的分布一致，则写出名称。
\item $F(y)=P(Y\le y)=P(X^2\le y)=$
\end{itemize}


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问：设随机变量 $X$ 的密度函数如下，
\begin{eqnarray*}
p(x)=\left\{\begin{array}{ll}
\frac{2x}{\pi^2}, & 0<x<\pi, \\
0, & \textrm{ 其它 } x.
\end{array}\right.
\end{eqnarray*}
求随机变量 $Y=\sin(X)$ 的密度函数。


\begin{itemize}
\item 先求分布函数，再求导得到密度函数。
\item $F(y)=P(Y\le y)=P[\sin(X)\le y]=$
\end{itemize}

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